Question

Design a grit chamber for a wastewater treatment plant with an average flowrate of 25,000 m3/d and a peak flowrate of 55,000 m/d. The detention time at peak flow is 3.0 min. The width to depth ratio is 2:1. Use a depth of 2 m. The aeration rate is 0.4 m/min per m of tank length. a) Determine the total air required, (15 points) b) Dimensions of the grit chamber, (15 points) c) Determine detention time at average flow. (10 points)

Answer 1

solution Given, Average flow rate (Q Aug. ) = 25,000 m3 day Peak flow rate (Speak) = 55,000 by Delention time at Peak Ita, Peak) = 3 min. W:D = 2:1 D = 2 m. Aeration rate (Ral = 0.4 m3 min perm of tank length Volume oben Detention time at peak Ltd, peak) = Speak volume of chamber (V) = td,peak Speak = 3 min x 55000 m3 day x24X60 min V - 114.583 m3 | volume of grit Chamber (V) - WXDXL 114.583 = (2x2)X2 XL L = 14.323 m RaXL. required = m 14.32 3 tande (a) Total air m3 = 0.4 length - X min. perm of tank length

= 5.7292 m² nain bien will, odmah mon in a day X 24X60 min. lotal air required - 57092 m3 min Total air required = 8250.05 ms 6 Dimensions of the grit chamber. Width = 4m Depth = 2 m length = 14.323 m. ya Volume of tank (C) Detention time at average flow Harg.) = more of coma, Saug. notilo lollol 114.583 m3 25000 day Plberde day - 114.583 2500 COT OYOYO 2 114.583 x2 4X60 min 20D and Itararg. = 6.6 min.

Additional of two tanks are used in grit chamber, then volume 52 of each tank (V.) = 2 . 114.583 . 02 Vez 57.29 is m3 rodoroch in V = WXDXL 57.2915 = (2X2)X2XL = 7. 16 14 m of we increase length by 20% to account for inlet & outlet conditions. Ilinni (06 Total length = L = 1.2x7.1614 Total length = 8.6 m. Then 0.48 8.68 24X60 m3 (a) Total air required = Total air required = 4953.6 m3 Ow4m D = 2m L = 8.6 m 6 Delentian time for average flow (taarg.) = 6.6 min.